您好我试图使用PHP将文件从我的Android应用程序上传到我的服务器。
我已阅读此帖:
How to upload a file using Java HttpClient library working with PHP http://www.veereshr.com/Java/Upload How do I send a file in Android from a mobile device to server using http?
这是我的JAVA代码:
public void upload() throws Exception {
File file = new File("data/data/com.tigo/databases/exercise");
Log.i("file.getName()", file.getName());
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost("http://***.***.***.***/backDatabase.php");
InputStreamEntity reqEntity = new InputStreamEntity( new FileInputStream(file), -1);
reqEntity.setContentType("binary/octet-stream");
reqEntity.setChunked(true);
httppost.setEntity(reqEntity);
HttpResponse response = httpclient.execute(httppost);
if((response.getStatusLine().toString()).equals("HTTP/1.1 200 OK")){
// Successfully Uploaded
Log.i("uploaded", response.getStatusLine().toString());
}
else{
// Did not upload. Add your logic here. Maybe you want to retry.
Log.i(" not uploaded", response.getStatusLine().toString());
}
httpclient.getConnectionManager().shutdown();
}
这是我的PHP代码:
<?php
$uploads_dir = '/tigo/databaseBackup';
if (is_uploaded_file($_FILES['exercise']['tmp_name']))
{
$info = "File ". $_FILES['exercise']['name'] ." uploaded successfully.\n";
$file = 'emailTest.log';
file_put_contents($file, $info, FILE_APPEND | LOCK_EX);
move_uploaded_file ($_FILES['exercise'] ['tmp_name'], $_FILES['exercise'] ['name']);
}
else
{
$info = "Possible file upload attack: ";
$file = 'emailTest.log';
file_put_contents($file, $info, FILE_APPEND | LOCK_EX);
$info = "filename '". $_FILES['exercise']['tmp_name'] . "'.";
file_put_contents($file, $info, FILE_APPEND | LOCK_EX);
print_r($_FILES);
}
?>
在我的logcat中,我得到HTTP / 1.1 200 OK。 当我查看服务器日志时,我收到此错误:
PHP Notice: Undefined index: exercise in /var/www/backDatabase.php on line 23
我也尝试过使用:
$_FILES['userfile']['name']
而不是
$_FILES['exercise']['tmp_name']
我的服务器日志中出现了同样的错误。
我认为我的问题是我无法参考我上传的文件。
感谢您的帮助。
答案 0 :(得分:4)
尝试多部分实体
public void upload(String filepath) throws IOException
{
HttpClient httpclient = new DefaultHttpClient();
httpclient.getParams().setParameter(CoreProtocolPNames.PROTOCOL_VERSION, HttpVersion.HTTP_1_1);
HttpPost httppost = new HttpPost("url");
File file = new File(filepath);
MultipartEntity mpEntity = new MultipartEntity();
ContentBody cbFile = new FileBody(file, "image/jpeg");
mpEntity.addPart("userfile", cbFile);
httppost.setEntity(mpEntity);
System.out.println("executing request " + httppost.getRequestLine());
HttpResponse response = httpclient.execute(httppost);
HttpEntity resEntity = response.getEntity();
// check the response and do what is required
}
答案 1 :(得分:0)
试试这个:
JAVA代码:
import java.io.File;
import org.apache.http.HttpEntity;
import org.apache.http.HttpResponse;
import org.apache.http.HttpVersion;
import org.apache.http.client.HttpClient;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.entity.mime.MultipartEntity;
import org.apache.http.entity.mime.content.ContentBody;
import org.apache.http.entity.mime.content.FileBody;
import org.apache.http.impl.client.DefaultHttpClient;
import org.apache.http.params.CoreProtocolPNames;
import org.apache.http.util.EntityUtils;
public class UploadFile {
public static void main(String[] args) throws Exception {
HttpClient httpclient = new DefaultHttpClient();
httpclient.getParams().setParameter(CoreProtocolPNames.PROTOCOL_VERSION, HttpVersion.HTTP_1_1);
HttpPost httppost = new HttpPost("http://***.***.***.***/backDatabase.php");
File file = new File("/data/data/com.tigo/databases/exercise");
MultipartEntity mpEntity = new MultipartEntity();
ContentBody cbFile = new FileBody(file);
mpEntity.addPart("userfile", cbFile);
httppost.setEntity(mpEntity);
System.out.println("executing request " + httppost.getRequestLine());
HttpResponse response = httpclient.execute(httppost);
HttpEntity resEntity = response.getEntity();
System.out.println(response.getStatusLine());
if (resEntity != null) {
System.out.println(EntityUtils.toString(resEntity));
}
if (resEntity != null) {
resEntity.consumeContent();
}
httpclient.getConnectionManager().shutdown();
}
}
PHP代码:
<?php
if (is_uploaded_file($_FILES['userfile']['tmp_name'])) {
echo "File ". $_FILES['userfile']['name'] ." uploaded successfully.\n";
move_uploaded_file ($_FILES['userfile'] ['tmp_name'], $_FILES['userfile'] ['name']);
} else {
echo "Possible file upload attack: ";
echo "filename '". $_FILES['userfile']['tmp_name'] . "'.";
print_r($_FILES); }
?>
您可能需要更改路径以满足您的需求,但上述代码可以正常工作。
答案 2 :(得分:-1)
这只是一小段代码。它可以将任何图像从Android发送到您的网站 使用Android的服务器
System.Net.WebClient Client = new System.Net.WebClient();
Client.Headers.Add("Content-Type", "binary/octet-stream");
byte[] result = Client.UploadFile("localhost/FolderName/upload.php", "POST", path);
string s = System.Text.Encoding.UTF8.GetString(result, 0, result.Length);
这是PHP代码{upload.php}。在中创建文件夹名称{Uploads} 你的申请。
<?php
$uploads_dir = 'uploads/'; //Directory to save the file that comes from client application.
if ($_FILES["file"]["error"] == UPLOAD_ERR_OK)
{
$tmp_name = $_FILES["file"]["tmp_name"];
$name = $_FILES["file"]["name"];
move_uploaded_file($tmp_name, "$uploads_dir/$name");
}
?>